Zipper（動態規劃）

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描述 Given three strings, you are to determine whether the third string can be formed by combining the characters in the first two strings. The first two strings can be mixed arbitrarily, but each must stay in its original order.

For example, consider forming "tcraete" from "cat" and "tree":

String A: cat
String B: tree
String C: tcraete

As you can see, we can form the third string by alternating characters from the two strings. As a second example, consider forming "catrtee" from "cat" and "tree":

String A: cat
String B: tree
String C: catrtee

Finally, notice that it is impossible to form "cttaree" from "cat" and "tree".
輸入 The first line of input contains a single positive integer from 1 through 1000. It represents the number of data sets to follow. The processing for each data set is identical. The data sets appear on the following lines, one data set per line.

For each data set, the line of input consists of three strings, separated by a single space. All strings are composed of upper and lower case letters only. The length of the third string is always the sum of the lengths of the first two strings. The first two strings will have lengths between 1 and 200 characters, inclusive.
輸出 For each data set, print:

Data set n: yes

if the third string can be formed from the first two, or

Data set n: no

if it cannot. Of course n should be replaced by the data set number. See the sample output below for an example.
樣例輸入

3
cat tree tcraete
cat tree catrtee
cat tree cttaree

樣例輸出

Data set 1: yes
Data set 2: yes
Data set 3: no

來源 Pacific Northwest 2004

花了好長時間終於用動態規劃的方法把樣例輸入的結果給弄對，但是提交的時候卻是WA，後來看了別人的代碼發現好多都是用記憶化搜索來做的

AC代碼

源代碼

#include <stdio.h>  
#include <string.h>  
  
int main()  
{  
    int  i, j, k, t;  
    int  L1, L2, ok[202][202];  
    char str1[201], str2[201], str3[402];  
                    /********************** 
                      ok[i][j]爲真時表示str1的前i個字符可以和str2的前j個字符 
                          組成str3的前i+j個字符。爲假時 表示不能。 
                    ***********************/  
  
    scanf("%d", &t);  
    for (i=1; i<=t; i++)  
    {  
        scanf("%s %s %s", str1, str2, str3);  
        memset(ok, 0, sizeof(ok));  
        L1 = strlen(str1);  
        L2 = strlen(str2);  
  
        ok[0][0] = 1;  
        for (j=1; j<=L1; j++)  
        {  
            if (ok[j-1][0] == 1 && str1[j-1] == str3[j-1])  
                ok[j][0] = 1;  
            else  
                break;  
        }  
        for (j=1; j<=L2; j++)  
        {  
            if (ok[0][j-1] == 1 && str2[j-1] == str3[j-1])  
                ok[0][j] = 1;  
            else  
                break;  
        }  
        for (j=1; j<=L1; j++)  
        {  
            for (k=1; k<=L2; k++)  
            {  
                if (ok[j-1][k]==1 && str1[j-1]==str3[j+k-1] || ok[j][k-1]==1 && str2[k-1]==str3[j+k-1])  
                    ok[j][k] = 1;  
            }  
        }  
        if (ok[L1][L2] == 1)  
            printf("Data set %d: yes\n", i);  
        else  
            printf("Data set %d: no\n", i);  
    }  
}

WA代碼

源代碼

#include<stdio.h>
#include<string.h>
#include<vector>
#include<algorithm>
#include<iostream>

using namespace std;
char strr1[1010];
char strr2[1010];
char strr3[1010];
int maxLen[1010][1010];
int main()
{
	int n;
	scanf("%d",&n);
	for(int w=1; w<=n; w++)
	{
		scanf("%s %s %s",&strr1,&strr2,&strr3);
		int length1=strlen(strr1);
		int length2=strlen(strr2);
		int length3=strlen(strr3);
		int nTmp;
//		int i,j,k;
		memset(maxLen,0,sizeof(maxLen));
		for(int i=0; i<=length1; i++)
		{
			maxLen[i][0]=0;
		}
		for(int k=0; k<=length3; k++)
		{
			maxLen[0][k]=0;
		}
		for(int i=1; i<=length1; i++)
		{
			for(int k=1; k<=length3; k++)
			{
				if(strr1[i-1]==strr3[k-1])
				maxLen[i][k]=maxLen[i-1][k-1]+1;
				else
				maxLen[i][k]=max(maxLen[i-1][k],maxLen[i][k-1]);
			}			
		}
		if(maxLen[length1][length3]==length1)
		{
			memset(maxLen,0,sizeof(maxLen));
		
	    	for(int j=0; j<length2; j++)
			{
				maxLen[j][0]=0;
			}			
			
			for(int k=0; k<length3; k++)
			{
				maxLen[0][k]=0;
			}			

			for(int j=1; j<=length2; j++)
				{
					for(int k=1; k<=length3; k++)
					{
						if(strr2[j-1]==strr3[k-1])
						maxLen[j][k]=maxLen[j-1][k-1]+1;
						else
						maxLen[j][k]=max(maxLen[j-1][k],maxLen[j][k-1]);
					}
				}
			if(maxLen[length2][length3]==length2)
			printf("Data set %d: yes\n",w);
			else
			printf("Data set %d: no\n",w);
			
		}
		else
		printf("Data set %d: no\n",w);	
	}
	
	return 0;
}